线性代数（最小割，最大密度子图，TJOI2015）

题意

给出一个 (N imes N) 的矩阵 (B) 和一个 (1 imes N) 的矩阵 (C)。

求出一个 (1 imes N) 的 (01) 矩阵 (A)，使得 (D = (A imes B−C）×A^T)最大。

输出 (D)。

思路

先对式子进行化简：
不妨设(B = (a_{ij})_{n imes n})，(C = (c_i)_n)

[egin{aligned} D &= A imes B imes A^T - C imes A^T \ &= sum_{i = 1}^nsum_{j = 1}^na_{ij}x_ix_j - sum_{i = 1}^nc_ix_i end{aligned} ]

发现这道题就是最大获利那道题的变形，考虑最大密度子图

两点之间的边权为(a_{ij} + a_{ji})，点权为(-c_i + a_{ii})

代码

#include <iostream>
#include <cstdio>
#include <cstring>
#include <algorithm>
#include <queue>

using namespace std;

const int N = 510, M = (250000 + N * 2) * 2, inf = 1e8;

int n, S, T;
int h[N], e[M], ne[M], f[M], idx;
int cur[N], d[N];
int dg[N], p[N];
int w[N][N];

void add(int a, int b, int c1, int c2)
{
    e[idx] = b, f[idx] = c1, ne[idx] = h[a], h[a] = idx ++;
    e[idx] = a, f[idx] = c2, ne[idx] = h[b], h[b] = idx ++;
}

bool bfs()
{
    memset(d, -1, sizeof(d));
    queue<int> que;
    que.push(S);
    d[S] = 0, cur[S] = h[S];
    while(que.size()) {
        int t = que.front();
        que.pop();
        for(int i = h[t]; ~i; i = ne[i]) {
            int ver = e[i];
            if(d[ver] == -1 && f[i]) {
                d[ver] = d[t] + 1;
                cur[ver] = h[ver];
                if(ver == T) return true;
                que.push(ver);
            }
        }
    }
    return false;
}

int find(int u, int limit)
{
    if(u == T) return limit;
    int flow = 0;
    for(int i = cur[u]; ~i && flow < limit; i = ne[i]) {
        cur[u] = i;
        int ver = e[i];
        if(d[ver] == d[u] + 1 && f[i]) {
            int t = find(ver, min(f[i], limit - flow));
            if(!t) d[ver] = -1;
            f[i] -= t, f[i ^ 1] += t, flow += t;
        }
    }
    return flow;
}

int dinic()
{
    int res = 0, flow;
    while(bfs()) {
        while(flow = find(S, inf)) {
            res += flow;
        }
    }
    return res;
}

int main()
{
    scanf("%d", &n);
    memset(h, -1, sizeof(h));
    S = 0, T = n + 1;
    for(int i = 1; i <= n; i ++) {
        for(int j = 1; j <= n; j ++) {
            int x;
            scanf("%d", &x);
            w[i][j] = x;
        }
    }
    for(int i = 1; i <= n; i ++) {
        for(int j = i + 1; j <= n; j ++) {
            add(i, j, w[i][j] + w[j][i], w[i][j] + w[j][i]);
            dg[i] += w[i][j] + w[j][i];
            dg[j] += w[i][j] + w[j][i];
        }
    }
    for(int i = 1; i <= n; i ++) {
        scanf("%d", &p[i]);
        p[i] = -p[i] + w[i][i];
    }
    int U = 0;
    for(int i = 1; i <= n; i ++) U = max(U, 2 * p[i] + dg[i]);
    for(int i = 1; i <= n; i ++) add(S, i, U, 0);
    for(int i = 1; i <= n; i ++) add(i, T, U - 2 * p[i] - dg[i], 0);
    printf("%d
", (n * U - dinic()) / 2);
    return 0;
}